Related Rates

Related rates problems deal with situations in
which several things are changing at rates which are
related. The way in which the rates are related often arises
from geometry, for example.

Example. The radius of a circle increases at
2 light-years per fortnight. At what rate is the area increasing when
the radius is 3 light-years?

The area of a circle is , where r is the radius.
Differentiate this equation with respect to t:

I want to know the rate of change of the area, which is .

The rate at which the radius increases is , so . I want to find when . Plug in:

The units should be square light-years per fortnight.

Notice that I don't plug into the equation . I differentiate first, then plug in.

In most related rates problems, you will perform the steps above:
Differentiate a starting equation with respect to t (time), then plug
in and answer the question. In most cases, the difficult part of the
problem is obtaining the starting equation.

Before I give some examples involving real-world situations, I'll do
an example which just illustrates the kind of mathematics that is
involved.

Example. x and y are functions of t, and are
related by

Find the rate at which x is changing when and
, if y is decreasing at 3 units per second.

I differentiate the given equation with respect to t. To save
writing, I'll use to denote and to
denote .

Here's how I got some of the terms.

is a product, so I applied the Product Rule to get
. Note that the derivative of y is , and by the Chain Rule, the derivative of is

I differentiated using the Chain Rule:

I differentiated using the Chain Rule:

I plug in , , and into (*). (Note that is negative, because I was
told that y is decreasing.)

Simplify and solve for :

Example. Calvin and Phoebe elope in a hot
air balloon, which rises at a constant rate of 3 meters per second.
Five seconds after they cast off, Phoebe's jilted suitor Bonzo
McTavish races up in his Porsche. He parks 50 meters from the launch
pad, and runs toward the pad at 2 meters per second. At what rate is
the distance between Bonzo and the balloon changing when the balloon
is 30 meters above the ground?

Here's a picture:

The triangle is a right triangle, and I've labelled the three sides
{\it with variables}. x is the distance from Bonzo to the launch pad,
y is the height of the balloon, and s is the distance from Bonzo to
the balloon. I use variables because the three sides are changing
with time: The balloon rises (y changes), Bonzo runs toward the
launch pad (x changes), and the distance between Bonzo and the
balloon (s) changes.

There is an obvious relationship among the variables --- Pythagoras'
theorem says

Differentiate with respect to t:

Cancel the 2's:

Now I plug in the numbers. Bonzo's velocity is --- negative, because the distance from Bonzo to the
launch pad is {\it decreasing}. is the rate at which the
balloon rises. I want to know what is when
. I see that I still need x and s.

The balloon was already 15 ( ) meters high when Bonzo drove up.
It has risen an additional meters. Since it's rising at
3 meters per second, additional seconds elapse.

During these 5 seconds, Bonzo is running at 2 meters per second, so
he travels meters. That leaves him meters from the launch pad. In other words,
.

Finally, using the equation ,

Plug all this stuff into the derivatives equation:

Here are some rules of thumb for setting up a related rates problem:

Draw a picture. Be sure you've drawn it at an arbitrary
instant. (In the previous example, it would be a bad idea to draw the
picture with the balloon on the ground, or with Bonzo just getting
out of his car.)

Label the quantities that are relevant to the problem. Quantities
that are changing should be labelled with {\it variables}.

Look for a relationship among the variables. Think of Pythagoras'
theorem, trigonometry, area or volume formulas, and similar
triangles.

Once you have a starting equation, differentiate with respect to t,
Then plug in the given values and solve for the appropriate rate.

In the next example, I'll need the following facts from physics. The
velocity of an object dropped from rest is feet per second after
t seconds. During that t seconds, it will have fallen
feet.

Example. A bowl of petunias is dropped from
a point 12 feet above the ground, so that it will land 4 feet from
the base of a 12 foot lamppost. At what rate is the shadow of the
bowl moving along the ground when the bowl is 8 feet above the
ground?

Here's a picture:

x is the distance from the shadow to the post. (In order to measure
the rate at which the shadow is moving, I have to measure its
distance to some fixed object.) h is the height of the bowl above the
ground.

By similar triangles,

(Notice that if you label the picture properly, it's usually obvious
what mathematical formula you should use.)

Clear denominators:

Differentiate with respect to t, being careful to use the Product
Rule on the right side:

Now I plug the numbers in. I'm trying to find , the rate at which the shadow is moving.

When the bowl is 8 feet above the ground, , so the bowl has
fallen 4 feet. By the physics fact I mentioned earlier, the bowl
falls feet in t seconds; setting , I get . The bowl has been falling for second.

By the other physics fact I mentioned, its velocity is feet per second. Thus, --- negative, because h is decreasing.

The only other thing I need is x. Set in :

Substitute x, h, and :

Example. Calvin and Bonzo drive on long
parallel roads in opposite directions at 20 feet per second and 80
feet per second, respectively. The roads are 200 feet apart. How
rapidly is the distance between them changing 5 seconds after they
pass one another?

Here's a picture:

x is the horizontal separation between Calvin and Bonzo. s is the
distance between them.

By Pythagoras,

Differentiate with respect to t:

I want .

The rate of change of the distance between them is the same as it
would be if Bonzo was moving at feet per second and
Calvin was stationary. So , and 5 second after they
pass each other, .

To find s, plug into :

Now plug everything in:

Example. A ferris wheel 50 feet in diameter
makes 2 revolutions per minute. Assume that the wheel is tangent to
the ground and let P be the point of tangency. At what rate is the
distance between P and a rider changing, when she is 25 feet above
the ground and going up?

Here's a picture:

is the angle between the vertical and a radius out to
the rider. s is the distance between the bottom of the wheel and the
rider.

By the Law of Cosines,

Differentiate with respect to t:

Assuming that the wheel is turning counterclockwise in the picture,
when the rider is 25 feet above the ground, she's on the far right
hand side. At this point, , and (since the triangle is then an isosceles right
triangle).

is the rate at which the wheel is turning. I must be
careful to express this in radians per minute:

Plug in:

Example. A spotlight on the top of a police
cruiser makes one revolution per second. The spotlight is 40 feet
from a long straight wall. At what rate is the spot of light moving
across the wall at the instant when the beam makes a angle with the wall?

Here's a picture:

x is the distance from the spot to the point on the wall nearest the
cruiser. is the angle between the beam and the altitude to the
wall from the light.

By basic trigonometry,

Differentiate with respect to t:

When the beam makes a angle with the wall, . The rate at which the beam is
turning is

Plug in:

Example. A helium balloon is released from
ground level and rises at a constant rate of 2 feet per second.
Calvin stands at the top of a 100-foot observation tower located 60
feet from the point where the balloon was released. At what rate is
the distance between Calvin and the balloon changing when the balloon
is 55 feet above the ground?

Let y be the height of the balloon above the ground, and let r be the
distance from Calvin to the balloon.